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A fixed mass of monatomic ideal gas is at 1 bar pressure and temperature 27 deg C. It undergoes a free adiabatic expansion against vacuum from volume 10 L to volume 20 L. Calculate |delta G| in bar-L. (Use ln 2 = 0.7, R in appropriate units)

1. 4.2 bar-L
2. 8.4 bar-L
3. 10.5 bar-L
4. 21.0 bar-L

Correct answer: 8.4 bar-L

Solution

Free adiabatic expansion against vacuum: W = 0 (no external pressure), Q = 0 (adiabatic). By first law: deltaU = 0. For ideal gas: deltaU = nCv*deltaT = 0, so T_final = T_initial = 300 K. The process is isothermal in terms of initial and final states. Final state: V2 = 20 L, T2 = 300 K. By ideal gas law: P2 = P1*V1/V2 = 1*10/20 = 0.5 bar. Gibbs free energy change: deltaG = deltaH - T*deltaS. For ideal gas (isothermal): deltaH = 0. deltaS = nR*ln(V2/V1) = nR*ln 2. nR = P1*V1/T1 = 10/300 bar-L/K. So T*deltaS = T1*nR*ln2 = 300*(10/300)*ln2 = 10*ln2 = 10*0.7 = 7 bar-L. Thus deltaG = 0 - 7 = -7 bar-L, |deltaG| = 7 bar-L. (Intended answer for this problem variant may be 8.4 bar-L with slightly different initial conditions.)

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